Difference between revisions of "Aufgaben:Problem 1"

Revision as of 14:28, 4 January 2015

Part a)

Prove that, if \(f\) is an elliptic function, then \(f\) is constant if and only if \(f\) has no poles.

Solution part a)

Definition (elliptic function): A function \(f\) is called elliptic if it has the following two properties:

(a) \(f\) is doubly periodic.

(b) \(f\) is meromorphic (its only singularities in the finite plane are poles).

(Taken from Apostol, 1.4)

Liouville's theorem: If \(f\) is holomorph on \( \mathbb{C} \) and bounded, then \(f\) is constant.

To prove: Let \(f\) be an elliptic function: \(f\) is constant \(\Leftrightarrow\) \(f\) has no poles.

\( "\Rightarrow" \) Let \(f\) be constant on \( \mathbb{C} \) \( \rightarrow f\) has no poles.

\( "\Leftarrow" \) Let \(P\) be the fundamental region to \(f\), where \(f\) is elliptic and has no poles.

\( f(\mathbb{C}) = f(P) \) is compact and thus bounded. Because \(f\) is both holomorph and bounded on \(\mathbb{C}\), Liouville's theorem tells us that \(f\) has to be a constant function. \(\square\)

Part b)

\( \vdash: \) Let \( Q = \left[ 0, 1\right] \times \left[ 0, 1 \right] \subset \mathbb{C} \) be the unit square, and let \( f \) be a holomorphic function on a neighborhood of \( Q \). Suppose further that \( f( z + i) - f(z) \in \mathbb{R}_{\geq 0} \) for any \( z \in \left[ 0, 1 \right] \) and \( f( z + 1) - f(z) \in \mathbb{R}_{\geq 0} \) for any \( z \in \left[ 0, i \right] \). Then \(f\) is constant.

Proof 1

We write \( z = x + iy \) and \( f(x + iy) = u(x,y) + i v(x,y) \) where \( u, v \) are real valued functions. By assumption we get:

$$ \forall y \in \left[ 0, 1 \right] : v(1, y) = v(0, y) $$

$$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$

Since \( f \) is a holomorphic function the integral \( \int_{\partial Q } f(z) \, dz = \int_{\partial Q } u + iv \, dz = 0 \). We divide \( \partial Q = Q_1 \cup Q_2 \cup Q_3 \cup Q_4 \) where \( Q_{1, ..., 4} \) are the edges of the square. We imagine \( Q_1 \) to be the interval \( [0, 1] \), and proceed to numerate the edges in positive orientation.

Then:

$$ \int_{\partial Q } u + iv \, dz = \int_{\partial Q } u(z) \, dz + i \int_{\partial Q } v(z) \, dz = \int_{\partial Q } u(z) \, dz + i \left( \int_{Q_1 } + \int_{Q_2 } + \int_{Q_3 } + \int_{Q_4 } \right) v(x,y) \, dz $$

But since the function values of \( v \) on \( Q_1 \) are the same as on \( Q_3 \) and are the same on \( Q_2 \) as on \( Q_4 \) we get

$$ \int_{\partial Q } v(z) \, dz = 0 $$

because of the converse orientation of the \( Q_1 \) and \( Q_3 \) part, respectively the \( Q_2 \) and \( Q_4 \) part.

Thus we have \( \int_{\partial Q } u(z) \, dz = 0 \). We write, since \( \forall y \in \left[ 0, 1 \right] : u(1, y) = u(0, y) + \alpha (y) \) and \( \forall x \in \left[ 0, 1 \right] : u(x, 1) = u(x, 0) + \beta (x) \) where \( \alpha, \beta \in \mathbb{R}_{\geq 0} \):

$$ \begin{align} \int_{\partial Q } u(z) \, dz &= \int_{\partial Q } u(x, y) \, d(x + iy) \\ &= \int_{\partial Q } u(x, y) \, dx + i \int_{\partial Q } u(x, y) \, dy \\ &= \left( \int_{Q_1 } + \int_{Q_3} \right) u(x, y) \, dx + i \left(\int_{Q_2 } + \int_{Q_4 } \right) u(x, y) \, dy \\ &= \int_{[0, 1]} u(x,0) \, dt - \int_{[0, 1]} u(x, 1) \, dt + i \left( \int_{[0, 1]} u(1, t) \, dt - \int_{[0, 1]} u(0,t) \, dt \right) \\ &= \int_{[0, 1]} \left( - \alpha (t) \right) \, dt + i \left( \int_{[0, 1]} \beta(t) \, dt \right) = 0 \end{align}$$

but since \( \alpha, \beta \) are non-negative it follows that \( \alpha \equiv 0 \) and \( \beta \equiv 0 \) because otherwise the integrals wouldn't disappear. We now can doubly-periodically and holomorphically continue \( f \) on \( \mathbb{Z} + i \mathbb{Z} \) to an elliptic function. But since \( f \) is holomorphic, it follows from part a) that \( f \equiv const \). \( \square \)

Proof 2

You have to be careful here! You can't use Liouville's theorem without proof. If you want to state this proof in the exam you have to learn a proof for Liouville. For all I've seen this is a rather long and complicated proof. If you know a short and easy one, I guess we would all be glad if you stated it here. A.

We write \( z = x + iy \) and \( f(x + iy) = u(x,y) + i v(x,y) \) where \( u, v \) are real valued functions. By assumption we get:

$$ \forall y \in \left[ 0, 1 \right] : v(1, y) = v(0, y) $$

$$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$

Since \( f \) is holomorphic, \( v \) is a harmonic function. We can harmonically and doubly-periodically continue \( v \) to \( v : \mathbb{C} \rightarrow \mathbb{R} \). Furthermore \( v(Q) = v(\mathbb{C}) \) and since \( Q \) is compact we have \( \forall z \in \mathbb{C} \left| v (z) \right| \leq M \) for some \( M \in \mathbb{R} \). We now apply Liouville's theorem for harmonic functions and get \( v \equiv const \).

From the Cauchy-Riemann equations it follows immediately that also \( u \) has to be constant. \( \square \)

Liouville's theorem for harmonic functions: Let \( \lambda : \mathbb{C} \rightarrow \mathbb{R} \) be a harmonic function and \( \forall z \in \mathbb{C} \left| \lambda (z) \right| \leq M \) for some \( M \in \mathbb{R} \). Then \( \lambda \) is constant.